A hovercraft of mass 83.0 kg can move on a horizontal surface, the x-y plane. A single unbalanced force acts on the hovercraft, but the size

Question

A hovercraft of mass 83.0 kg can move on a horizontal surface, the x-y plane. A single unbalanced force acts on the hovercraft, but the size of the force is unknown. The hovercraft initially has a velocity of 4.0 m/s in the positive x direction and some time later has a velocity of 7.0 m/s in the positive y direction. How much work is done on the hovercraft by the force during this time?

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RI SƠ 4 years 2021-08-29T00:30:00+00:00 1 Answers 9 views 0

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  1. phucdien
    0
    2021-08-29T00:31:36+00:00
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    Answer:

    1369.5 J

    Explanation:

    According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy of the object itself.

    Therefore, we can write:

    W=K_f-K_i=\frac{1}{2}mv^2-\frac{1}{2}mu^2

    where:

    W is the work done on the hovercraft

    m = 83.0 kg is the mass of the hovercraft

    u = 4.0 m/s is the initial velocity (in the positive x-direction)

    v = 7.0 m/s is the final velocity (in the positive y-direction)

    Here the direction of motion of the hovercraft has changed; however, we can ignore this fact, since work is a scalar so the direction of the velocity does not matter.

    Therefore, the work done by the force is:

    W=\frac{1}{2}(83.0)(7.0)^2-\frac{1}{2}(83.0)(4.0)^2=1369.5 J

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